Molarity Of Sodium Chloride Solution

To make a 15% w/v sodium chloride solution, we need to determine the amount of sodium chloride (in grams) needed in 200 ml of solution. The percentage w/v means that we have 15 grams of sodium chloride in 100 ml of solution. Therefore, to find the amount needed in 200 ml, we can set up a proportion: 15g/100ml = x/200ml. Cross-multiplying gives us 100x = 15 * 200, which simplifies to 100x = 3000. Dividing both sides by 100, we find that x = 30g. Therefore, 30g of sodium chloride is needed to make a 200 ml, 15% w/v sodium chloride solution.

In this experiment, a serial dilution of serum is made by transferring 1ml of serum into tube #1 and adding 1ml of saline. Then, 1ml of the dilution from tube #1 is transferred into tube #2 and mixed with 1ml of saline. This process is repeated until tube #10. The tubes are then tested for the presence of anti-HIV antibodies. The fact that tubes #1 to #7 tested positive suggests that the dilution in these tubes was not strong enough to eliminate the presence of antibodies. Tube #7, which showed the highest dilution with a positive reaction, indicates that the antibodies were still present at this dilution. Therefore, the dilution in tube #7 must be the highest among the positive tubes, which is 1/128.

To make a 15% w/v sodium chloride solution, 15g of sodium chloride is needed. This means that for every 100ml of the solution, there should be 15g of sodium chloride. Since we want to make 1000ml of the solution, we need to multiply 15g by 10, resulting in 150g of sodium chloride needed.

To make a 15% w/v sodium chloride solution, we need 15g of sodium chloride. This means that for every 100ml of solution, 15g of sodium chloride should be dissolved in it. The "w/v" stands for weight/volume, indicating that the concentration is expressed as a percentage of weight per volume. Therefore, the correct answer is 15g.

The titer refers to the highest dilution of a substance that still produces a positive reaction. In this case, the first 5 tubes tested positive, indicating that the highest dilution that still showed a positive reaction was tube #5. Since each tube represents a 1:10 dilution, the titer can be calculated by multiplying the dilution factor of the last positive tube (tube #5) by the dilution factor of each tube (1:10). Therefore, the titer is 10^5, which is equal to 32.

Making a really big dilution in smaller 10 fold compound dilutions is more accurate because it allows for better control and precision in the dilution process. By gradually diluting the compound in smaller increments, any errors or inconsistencies can be easily identified and corrected. On the other hand, making a large dilution in one step can lead to larger errors and make it more difficult to accurately determine the concentration of the resulting solution.

The dilution on tube #5 is 1/32. This is because each tube in the serial dilution was diluted by a factor of 1/2. Starting with tube #1, which contained 1ml of serum, each subsequent tube had 1ml of the previous dilution added to it. Therefore, the dilution factor increased by 1/2 with each tube. Tube #5 would have undergone a dilution of 1/2 * 1/2 * 1/2 * 1/2 * 1/2 = 1/32.

The titer is determined by the highest dilution that still shows a positive reaction. In this case, tube #7 is the highest dilution that tested positive. Since each dilution is made by a factor of 2 (1:2 dilution), we can calculate the titer by multiplying the dilution factor of each tube starting from tube #1. Tube #1 has a dilution factor of 1, tube #2 has a dilution factor of 2, tube #3 has a dilution factor of 4, and so on. Therefore, the dilution factor for tube #7 is 2^7 = 128, which means the titer is 128.

The best way to make a 10mL, 1/1000 reagent dilution is to make a 10-fold compound dilution with each tube 1/10 as concentrated as the last tube. Three tubes are needed - tube #1, #2, and #3. Tube #1 contains 1ml reagent and 9ml diluent. Tube #2 contains 1ml of dilution from tube #1 and 9ml diluent. Tube #3 contains 1ml of dilution from tube #2 and 9ml diluent. Tube #3 now contains 10mL of 1/1000 dilution of reagent.

To make a 5M, 5L sodium hydroxide solution, the amount of sodium hydroxide needed can be calculated using the formula:
moles = concentration (M) x volume (L).
Since the concentration is 5M and the volume is 5L, the moles of sodium hydroxide required is 5 x 5 = 25 moles.
The molar mass of sodium hydroxide (NaOH) is approximately 40 g/mol.
Therefore, the mass of sodium hydroxide needed can be calculated by multiplying the moles by the molar mass:
25 moles x 40 g/mol = 1000g.
Hence, 1000g of sodium hydroxide is needed to make a 5M, 5L sodium hydroxide solution.

To calculate the amount of sodium chloride needed to make a 3M, 10L solution, we need to use the formula: Molarity (M) = moles of solute / liters of solution. Rearranging the formula, moles of solute = Molarity (M) * liters of solution. In this case, the molarity is 3M and the volume is 10L. Plugging in these values, we get moles of solute = 3M * 10L = 30 moles. The molar mass of sodium chloride (NaCl) is 58.44 g/mol. Multiplying the moles by the molar mass, we get 30 moles * 58.44 g/mol = 1740g of sodium chloride needed.

To make a 3M sodium chloride solution, we need to calculate the number of moles of NaCl required. The molar mass of NaCl is 58.44 g/mol.

First, we need to convert the volume from mL to L by dividing by 1000: 10 mL ÷ 1000 = 0.01 L.

Next, we use the formula Molarity (M) = moles of solute ÷ volume of solution in liters. Rearranging the formula, moles of solute = Molarity × volume of solution in liters.

Substituting the given values, moles of solute = 3M × 0.01 L = 0.03 moles.

Finally, we use the formula moles = mass ÷ molar mass to calculate the mass of NaCl. Rearranging the formula, mass = moles × molar mass.

Substituting the values, mass = 0.03 moles × 58.44 g/mol = 1.7532 g, which rounds to 1.74 g. Therefore, the correct answer is 1.74g.

To make a 5M sodium hydroxide solution, we need to determine the amount of NaOH required. The molar mass of NaOH is approximately 40 g/mol. Since we want a 5M solution, we need 5 moles of NaOH in 1 liter of solution. As we only need a 5mL solution, we can calculate the amount of NaOH needed by converting the volume to liters and multiplying it by the molarity. Therefore, 5 mL is equivalent to 0.005 liters. Multiplying 0.005 liters by 5 moles gives us 0.025 moles of NaOH. To convert moles to grams, we multiply by the molar mass, resulting in 1g of NaOH needed for the solution.

The correct answer suggests making a 10 fold compound dilution with each tube 1/10 as concentrated as the last tube. This means that in each tube, the concentration of the reagent is diluted by a factor of 10. Two tubes are needed - #1 and #2. In tube #1, 1ml of reagent is mixed with 9ml of diluent, resulting in a 10-fold dilution. Then, in tube #2, 1ml of the dilution from tube #1 is mixed with 9ml of diluent, resulting in another 10-fold dilution. Therefore, tube #2 now contains a 1/100 dilution of the reagent in a total volume of 10mL.

To make a 15% w/v sodium chloride solution, we need 150g of sodium chloride. This means that for every 100mL of solution, there should be 15g of sodium chloride. Since we want to make 1L of solution, we need to multiply the 15g by 10, resulting in 150g of sodium chloride needed.