MCQ
The mole fraction of a solute in a $100$ molal aqueous solution .......... $\times 10^{-2}$
(Round off to the Nearest Integer).
[Given : Atomic masses : $H : 1.0 \,u , O : 16.0\, u ]$
- ✓$64$
- B$52$
- C$44$
- D$62$
(Round off to the Nearest Integer).
[Given : Atomic masses : $H : 1.0 \,u , O : 16.0\, u ]$
mole-fraction of solute $=\frac{ n _{\text {solute }}}{ n _{\text {solute }}+ n _{\text {solvent }}}$
$=\frac{100}{100+\frac{1000}{18}}=\frac{1800}{2800}=0.6428$
$=64.28 \times 10^{-2}$
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$K_1 = 4.2 \times 10^{-7}$ and $K_2 = 4.8 \times 10^{-11}$
Select the correct statement for a saturated $0.034\, M$ solution of the carbonic acid.
