MCQ
For $0.1\,M $ solution, the colligative property will follow the order
- A$NaCl > N{a_2}S{O_4} > N{a_3}P{O_4}$
- ✓$NaCl < N{a_2}S{O_4} < N{a_3}P{O_4}$
- C$NaCl > N{a_2}S{O_4}\,\, \approx \,\,N{a_3}P{O_4}$
- D$NaCl < N{a_2}S{O_4} = N{a_3}P{O_4}$
$N{a_3}P{O_4} > N{a_2}S{O_4} > NaCl$
$N{a_3}P{O_4} \to 3N{a^ + } + PO_4^{3 - } = 4$
$N{a_2}S{O_4} \to 2N{a^ + } + SO_4^{2 - } = 3$
$NaCl \to N{a^ + } + C{l^ - } = 2$
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Statments $I:$ Maltose has two $\alpha$-D-glucose units linked at $C _{1}$ and $C _{4}$ and is a reducing sugar.
Statement $II:$ Maltose has two monosaccharides: $\alpha$-D-glucose and $\beta$-D-glucose linked at $C_{1}$ and $C_{6}$ and it is a non-reducing sugar.
In the light of the above statements, choose the correct answer from the options given below.
| $1$ | $2$ | $3$ | |
| $\mathrm{HI}\left(\mathrm{mol} \mathrm{L}^{-1}\right)$ | $0.005$ | $0.01$ | $0.02$ |
| Rate $\left(\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}-1\right)$ | $7.5 \times 10^{-4}$ | $3.0 \times 10^{-3}$ | $1.2 \times 10^{-2}$ |
$Pt|H_2(P_1\ atm)\ | \ H^+(x_1M)\ ||\ H^+(x_2M)\ |\ H_2(P_2\ atm)\ |Pt$ .
The cell reaction will be spontaneous if