MCQ
When $CuSO_4$ is added to a solution of ammonia
  • A
    Freezing point is lowered
  • Freezing point is raised
  • C
    Boiling point is raised
  • D
    Both $(A)$ and $(C)$

Answer

Correct option: B.
Freezing point is raised
b
Copper sulphate in ammonia solution form a complex which will lead to decrease the number of particles in solution. So, van't hoff factor will decrease.

Depression in freezing point and elevation in boiling point are directly proportional to the van 't hoff factor. Decreases in number of particles causes the increase in freezing point and decrease in the boiling point.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

The primary and secondary valencies of cobalt respectively in $\left[ Co \left( NH _3\right)_5 Cl \right] Cl _2$ are :
A solution of phenol in chloroform when treated with aqueous $NaOH$ gives compound $P$ as a major product. The mass percentage of carbon in $P$ is..............(to the nearest integer)

(Atomic mass : $C =12 ; H =1 ; O =16$ )

The laboratory test 'tailing the mercury' is applied to identify
Reagent used for detection of $Ni^{2+}$ is $A. x$ mol of $A$ reacts with $Ni^{2+}$ and give a complex of $C$ colour. $A, x, C$ are respectively
Which of the following elements is a lanthanide (Rare–earth element)
Which class of drugs are used to relieve stress?
Which of the following order of $CFSE$ is incorrect ?
Assertion : Acetoacetic ester,

$\begin{array}{*{20}{c}}
{O\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,O}\\
{||\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,||}\\
{C{H_3} - C - C{H_2}CO{C_2}{H_5}}
\end{array}$ will give iodoform test

Reason : It does not contains $\begin{array}{*{20}{c}}
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,O\,\,}\\
{\,\,\,\,{\kern 1pt} \,\,\,\,\,\,\,\,\,\,\,||\,\,\,}\\
{C{H_3} - C - }
\end{array}$ group

Among the given pairs, in which pair second compound has less enol content than first compound
The correct option($s$) about entropy ($s$) is(are)

$[ R =$ gas constant, $F =$ Faraday constant, $T =$ Temperature $]$

$(A)$ For the reaction, $M (s)+2 H ^{+}(a q) \rightarrow H _2(g)+ M ^{2+}(a q)$, if $\frac{ dE _{c o l l}}{ dT }=\frac{ R }{ F }$, then the entropy change of the reaction is $R$ (assume that entropy and internal energy changes are temperature independent).

$(B)$ The cell reaction, $Pt (s) \mid H _2(g, 1$ bar $)\left| H ^{+}(a q, 0.01 M ) \| H ^{+}(a q, 0.1 M )\right| H _2(g, 1 bar ) \mid Pt (s)$, is an entropy driven process.

$(C)$ For racemization of an optically active compound, $\Delta S >0$.

$(D)$ $\Delta S >0$, for $\left[ Ni \left( H _2 O \right)_6\right]^{2+}+3$ en $\rightarrow\left[ Ni ( en )_3\right]^{2+}+6 H _2 O$ (where en $=$ ethylenediamine).