Question

Prepare a 0.1N working standard solution in 250ml distilled water from a stock standard solution of 20% Na2CO3 in 50ml of distilled water. If you know that the Atomic Weight of Na=32, O=16, C=12:

a) 5 ml
b) 8.5 ml
c) 5.8 ml
d) 3 ml

Answer 1

To prepare a 0.1N sodium carbonate solution using the given stock solution, the calculation based on molar mass and equivalent mass leads to an estimated volume requirement of 0.1325 mL from the stock, but this does not match the provided options. The closest matching option to the calculated requirement is 5 mL.

Step-by-step explanation:

The question asks to prepare a 0.1N (normal) working standard solution of sodium carbonate (Na2CO3) using a 20% stock standard solution. First, the molar mass of Na2CO3 must be calculated using the atomic weights provided (Na=32, O=16, C=12), which is (2*32)+(3*16)+(12) = 106 g/mol. To find the equivalent mass, we divide the molar mass by the valency factor (which is 2 for Na2CO3), obtaining 53 g/equiv. Next, we calculate the amount of substance in the stock solution: 20% of 50 ml is 10 g of Na2CO3, which is equivalent to 10 g / 53 g/equiv = 0.1887 equiv. To prepare 250 ml of a 0.1N solution, we need 250 ml * 0.1N = 0.025 equiv. Therefore, we need to dilute 0.025 equiv / 0.1887 equiv/mL of stock solution, which yields approximately 0.1325 mL of stock is required.

However, the given options do not exactly match this calculation, indicating there might be an error in the question or a misunderstanding of the problem. Among the given options, 5 mL (option a) would be closest to the calculated requirement, assuming the calculated value is significantly lower due to an error.