Question

Dear sir /ma'am,

The solubility of sodium fluoride, NaF, in water at 25 degree Celsius is 4.100g per 100.0g of water. Assume 100.0g of water occupies a volume of 100.0ml.A saturated solution is prepared by adding 6.000gof NaF to 100.0ml of water at 25 degree Celsius in a beaker and steering thoroughly.

The following sequence of steps is carried out :

Step 1:50.0ml of solution is removed from the beaker and discarded.

Step 2:50.0ml of water is added to the beaker and steered.

Step 3:50.0ml of solution is Removed from the beaker and discarded.

The maximum mass of NaF now remaining and recoverable from the beaker is?

Options:

A) 1.025g

B) 1.900g

C) 1.975g

D) 4.100g

Please help me to solve it.

Answer 1

The solubility of NaF=4.100g per 100.0g of water.
Assume 100.0g of water occupies a volume of 100.0ml.
So, density of water=1g/cc.

So, 1 ml removed=1g removed.

A saturated solution is prepared by adding 6.000g of NaF to 100.0ml of water.
But 100 ml water can hold only 4.1 gram( solubility is 4.1)
So, rest 1.9 g is excess here, and this excess remains as solid.

The following sequence of steps is carried out :

Step 1:

50.0ml of solution is removed from the beaker and discarded.

We have 4.1g NaF dissolved in 100ml, and undissolved 1.9g salt.

So, in 50 ml, we will have 4.1/2=2.05 g

So, 2.05 g NaF is removed.
Balance, we have 50 ml solution, with 2.05 g dissolved NaF and 1.9g undissolved NaF.

Step 2:
50.0ml of water is added to the beaker and steered.
When 50 ml water is added again, NaF that was not dissolved earlier dissolves now.
So, in 50 ml, we can dissolve 2.05 g. But we have only 1.9 g undissolved salt. So, it completely dissolves.

So, we have 100 g solution , with 2.05+1.9=3.95 g salt.

Step 3:
50.0ml of solution is Removed from the beaker and discarded.
So, remaining is 50 ml.
So, balance salt= 3.95/2 = 1.975g.

So, maximum recoverable salt=1.975 g.