Question

An aqueous solution of glucose is one molal.
Amount of glucose (in grams) present in $1kg$ solution of glucose is :
$\text{Molar mass of glucose is}180gmo{l}^{-1}$

• A. $67.16g$
• B. $90g$
• C. $236.44g$
• D. $152.54g$

Answer 1

The correct option is D $152.54g$

Glucose solution is one molal.
Thus, $1000$ g water (solvent) has glucose = one mole = $180$ g
$\text{Total mass of solution}=(1000+180)g=1180g$
$\text{1180 g of solution has glucose}=180g$
$\therefore 1000g\text{of solution has glucose}=\frac{180\times 1000}{1180}=152.54g$