Match the list (Relative atomic mass : Na = 23 , K = 39 , Ca = 40 ) : List - I List - II (I) 20% ( w / w ) solution of KOH (density of solution = 1.02 gm / mL ) (P) 4.00 M (II) Solution containing 444

Question

Match the list (Relative atomic mass : Na = 23 , K = 39 , Ca = 40 ) : List - I List - II (I) 20% ( w / w ) solution of KOH (density of solution = 1.02 gm / mL ) (P) 4.00 M (II) Solution containing 444 gm CaCl 2 ​ in a 1 L solution (Q) 3.64 M (III) Volume of 1.204 × 1 0 24 molecules of water at 4 ∘ C (R) 5 mL (IV) Volume of 0.2 M NaOH solution containing 40 mg of NaOH (S) 36 mL The correct match is :

TestHub · Accepted Answer

(P)

20 gm of KOH present in 100 gm solution

$V = \frac{100}{1.02} = 98.04 ml$

$M = \frac{20/56}{98.04} \times 1000 = 3.64 M$

(Q)

444 gm CaCl 2 in 1 lit. solute

Molarity $= \frac{444}{111/1} = \frac{444}{111} = 4 M$

(R) $1.204 \times 1 0^{24}$ molecules of water $= 2$ mole

2 mole of water at $4^{\circ} C$ is 36 gm or 36 ml

(S)

Volume of 0.2 M NaOH solution

$0.2 \times V = 1 \times 1 0^{- 3}$ mole

$V = \frac{1 \times 1 0 ^{- 3}}{0.2} \times 1000$

$V = 5 ml$

Chemistry - Concentration Terms Question with Solution | TestHub

Options:

Doubts & Discussion

List - I	List - II
(I) $20% (w / w)$ solution of KOH (density of solution $= 1.02 gm / mL)$	(P) 4.00 M
(II) Solution containing $444 gm CaCl_{2}$ in a 1 L solution	(Q) 3.64 M
(III) Volume of $1.204 \times 1 0^{24}$ molecules of water at $4^{\circ} C$	(R) 5 mL
(IV) Volume of 0.2 M NaOH solution containing 40 mg of NaOH	(S) 36 mL