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Byju's Answer

Standard XII

Physics

Basic Differentiation Rule

Match the col...

Question

Match the column:
Column I Column II
(A) $20$ V $H_{2} O_{2}$ (P) $1.78$ M
(B) $17.45 %$ w/v $H_{2} S O_{4} (d = 1.1745$ g/ml) (Q) $1.78 m$
(C) Pure water (R) $1.5$ M
(D) $5 %$ w/w NaOH ( $d = 1.2 g m / m l$ ) (S) $55.5$ M

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Solution

(i0 For pure water the density is $1000 g / m l$
Mass of $H_{2} O = 18.2$
Molarity= $\frac{1000}{18.2} = 55.5 M$
(ii) $5$ % w/w $N a O H$
Molarity= $5 \times \frac{1.2}{40} = 1.5 M$
(A) $20 V H_{2} O_{2} ⟶ 1.78 m$
(B) $17.45$ % w/v $H_{2} S O_{4} ⟶ 1.78 M$
(C) Pure water $⟶ 55.5 M$
(D) $5$ % w/w $N a O H ⟶ 1.5 M$

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Column I	Column II
(A) $20$ V $H_{2} O_{2}$	(P) $1.78$ M
(B) $17.45 %$ w/v $H_{2} S O_{4} (d = 1.1745$ g/ml)	(Q) $1.78 m$
(C) Pure water	(R) $1.5$ M
(D) $5 %$ w/w NaOH ( $d = 1.2 g m / m l$ )	(S) $55.5$ M

Match the column:Column IColumn II(A) 20 V H2O2(P) 1.78M(B) 17.45% w/v H2SO4(d=1.1745 g/ml)(Q) 1.78m(C) Pure water(R) 1.5M(D) 5% w/w NaOH (d=1.2gm/ml)(S) 55.5 M

(i0 For pure water the density is 1000g/mlMass of H2O=18.2Molarity= 100018.2=55.5M(ii) 5 % w/w NaOHMolarity= 5×1.240=1.5M(A) 20VH2O2⟶1.78m(B) 17.45% w/v H2SO4⟶1.78M(C) Pure water⟶55.5M(D) 5% w/w NaOH⟶1.5M

Match the column:
Column I Column II
(A) $20$ V $H_{2} O_{2}$ (P) $1.78$ M
(B) $17.45 %$ w/v $H_{2} S O_{4} (d = 1.1745$ g/ml) (Q) $1.78 m$
(C) Pure water (R) $1.5$ M
(D) $5 %$ w/w NaOH ( $d = 1.2 g m / m l$ ) (S) $55.5$ M