Question

The teacher instructed three students A,B and C respectively to prepare a 50% ( mass by volume ) solution of sodium hydroxide. A dissolved 50 gram of NaOH in 100 ml of water, B dissolved 50 gram of NaOH in 100 gram of water while C dissolved 50 gram of NaOH in water to make 100 ml of solution.
a) Which one of the following has made the desired solution?

Answer 1

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Student A and B prepare 150 ml solution, so student C make desire solution because he adds water to make 100 ml solution. and from the calculation,

% w/v = 100 × weight of sub (solute)

volume of solution

∴ %50 = 100 × weight of sub / 100 ml

weight of sub = 50 ×100 ml / 100

∴ weight of sub = 50 g

Here the 50g NaOH required for 50% w/v 100ml solution of NaOH.

Answer 2

Answer:

Student C dissolved NaOH in water to make 100 mL of solution. He gets the desired 50% (mass by volume) solution of NaOH.

Explanation:

mass by volume percent of the desired solution is 50%.

the formula to find mass by volume percent of a solution is given by,

            $mass\; by\; volume\% = \frac{mass\;of \;solute}{volume \;of \:solution}\times100$

If the mass of solute is 50g

$volume\: of\; solution = \frac{mass \: of \; soulte}{mass\; by\;volume\%}\times100=\frac{50}{50}\times100=100mL$

we need 100mL solution by dissolving 50g of solute(NaOH) in water to make 50% (w/v) solution of NaOH.

Student  A took 50g of NaOH and 100mL of solvent.

Student B took 100g of water to dissolve 50g NaOH. since the density of water is 1g/mL then the volume of solvent is 100mL. Thus in both cases, the volume of the solution increases.

Student C dissolved 50g NaOH in water and then made up to 100ml. So, he gets the desired solution.