Q. The molality of a urea solution in which 0.0100 g of urea, $\lbrack(NH_{2}0_{2}CO\rbrack$is added to $0.3000\ {dm}^{3}$ of water at STP is
A) 0.555m
B) ${5.55 \times 10}^{- 4}$m
C) 33.3m
D) ${3.33 \times 10}^{- 2}$m
Explanation
(b)
Molality =$\frac{mole\ of\ solute}{kg\ of\ water}$
Moles of urea = $\frac{0.010}{60}$ mol
Water at STP ($d\ 1g/{cm}^{3} = 1kg/d{cm}^{3}$) = $0.3\ d{cm}^{3}$ = 0.3 kg
$\therefore\ Molality\ = \frac{0.010}{60\ \times 0.3} = 5.55 \times \ 10^{- 4}\ molal$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.