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प्रश्न
10.0 grams of caustic soda when dissolved in 250 cm3 of water, the resultant gram molarity of solution is _______.
(A) 0.25 M
(B) 0.5 M
(C) 1.0 M
(D) 0.1 M
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उत्तर
(C) 1.0 M [1]
Molarity of NaOH solution = `10/(40xx0.25=1.0M)`
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संबंधित प्रश्न
State Kohlrausch Law.
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| Concentration/M | 0.001 | 0.010 | 0.020 | 0.050 | 0.100 |
| 102 × κ/S m−1 | 1.237 | 11.85 | 23.15 | 55.53 | 106.74 |
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(iii) \[\ce{Λ^0_{(HNO_3)} + \ce{Λ^0_m_{(NaOH)} - \ce{Λ^0_m_{(NaNO_3)}}}}\]
(iv) \[\ce{Λ^0_m_{(NH_4OH)} + \ce{Λ^0_m_{(HCl)} - \ce{Λ^0_m_{(NH_4Cl)}}}}\]
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\[\begin{array}{cc}
\end{array}\]\[\begin{bmatrix}
\ce{\Lambda^{\circ}_{H^+} = 350 S cm^2 mol^{-1}}\\
\ce{\Lambda^{\circ}_{CH_3COO^-} = 50 S cm^2 mol^{-1}}
\end{bmatrix}\]
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| S. No. | Conc. (M) |
k S cm−1 | ∧m S cm2 mol−1 |
| 1. | 1.00 | 111.3 × 10−3 | 111.3 |
| 2. | 0.10 | 12.9 × 10−3 | 129.0 |
| 3. | 0.01 | 1.41 × 10−3 | 141.0 |
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OR
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