Wline-ext400 ml of 0-1 M - HCl - extsolution - - Whhine IV -text200 ml - 24-5 - 1 text w - v - H - 2 SO - 4 - S -SO- 4 -2- text0-06 M illhline- T - Cl - text0-2 M Whlinelend array A- I - P - T - II - R - III - T - U - IV - SB- I - T - II - R - III - S - IV - P - QC- I - P - T - II - T - III - P - S - IV - QD- I - Q - P - II - T - III - T - U - IV - S

The correct option is C I P, T; II T; III P, S; IV Q(I) Molarity of cation =M_1V_1 + M_2V_2V_1+V_2 = 0.2 × 100 + 0.1 × 400500 = 0.65=0.12 Molarit of ...

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Byju's Answer

Standard XII

Chemistry

Molarity

Wline&+ext400...

Question

$\begin{matrix} List-I & List-II (I) & 100 ml of 0.2 M A l C l_{3} solution & (P) & Total concentration of cation(s)=0.12M + 400 ml of 0.1 M H C l solution (I I) & 50 ml of 0.4 M K C l + 50 ml H_{2} O & (Q) & [S O_{4}^{2 -}] = 2.5 M (I I I) & 30 ml of 0.2 M K_{2} S O_{4} + 70 ml H_{2} O & (R) & [C l^{-} =] 2.5 M (I V) & 200 ml 24.5 % (w/v) H_{2} S O_{4} & (S) & [S O_{4}^{2 -} =] 0.06 M (T) & [C l^{-} =] 0.2 M (U) & Total concentration of cation(s) = 0.2 M \end{matrix}$

I - P, T; I I - R; I I I - T, U; I V - S

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I - P, T; I I - T; I I I - P, S; I V - Q

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I - T; I I - R; I I I - S; I V - P, Q

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I - Q, P; I I - T; I I I - T, U; I V - S

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Solution

The correct option is B $I - P, T; I I - T; I I I - P, S; I V - Q$
(I) Molarity of cation $= \frac{M_{1} V_{1} + M_{2} V_{2}}{V_{1} + V_{2}} = \frac{0.2 \times 100 + 0.1 \times 400}{500} = \frac{0.6}{5} = 0.12$
Molarit of $C l^{-} = \frac{3 (0.2) 100 + 0.1 \times 400}{500} = 0.2$

(II) Molarity of cation $= \frac{50 \times 0.4 + 0}{100} = 0.2$
Molarity of $C l^{-} = \frac{0.4 \times 50 + 0}{100} = 0.2$

(III) Molarity of cation $= \frac{2 (0.2) 30 + 0}{100} = 0.12$
Molarity fo $S O_{4}^{2 -} = \frac{30 \times 0.2}{100} = 0.06$

(IV) $24.5 g H_{2} S O_{4} in 100 ml solution$
Molarity $= \frac{\frac{25.4}{98}}{0.1} = 2.5$
$∴$ Concentration of cation $= 2 \times 2.5 M$
Concentration of $S O_{4}^{2 -} = 2.5 M$

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