# IMAT 2011 Q68 [Sodium Chloride]

Sodium chloride (relative molecular mass = 58.5) has a solubility of 36.0 g per 100 g of water.
The density of the solution is 1.13 g/ml. Which of the following calculations would give the solubility in moles per litre?

A. 36.0 x 10/58.5
B. 36.0 x 1000/(58.5 x 1.13)
C. 36.0 x 10 x 1.13/58.5
D. 36.0 x 10 x 1.13/(58.5 x 136)
E. 36.0 x 1000 x 1.13/(58.5 x 136)

The first step is to convert sodium chloride into moles using the grams and Mw given: {\frac{36} {58.5}} moles of NaCl. Then we have to calculate the volume using the density equation; {{d=}\frac{m}{V}} which gives us {{V=}\frac{m} {d} = \frac{136 (total volume of solution)} {1.13mL}}. Then we convert this volume into litres by multiplying by 1000 and the final equation is {\frac {n} {V} = {[36 \cdot 1.13]} \cdot [ \frac {1000}{ 58.5} \cdot} 136], so the answer E.