A) 1000
B) 990
C) 9990
D) 90
Correct Answer: B
Solution :
Given, volume of\[N{{a}_{2}}C{{O}_{3}}\]solution = 250 mL mass of\[N{{a}_{2}}C{{O}_{3}}\]solution = 2.65 g molecular mass of\[N{{a}_{2}}C{{O}_{3}}=106\text{ }g\] \[\Rightarrow \] \[w=\frac{ENV}{1000}\] Or \[2.65=\frac{\frac{106}{2}\times N\times 250}{1000}\] (\[\because \] eq wt of\[N{{a}_{2}}C{{O}_{3}}=\frac{106}{2}\]) \[N=0.2\] \[\because \]10 mL of this solution is added to\[x\]mL of water to obtain\[0.001\text{ }M\text{ }N{{a}_{2}}C{{O}_{3}}\]solution, therefore volume of solution taken, \[{{\text{V}}_{1}}=10\text{ }mL\] normality of solution,\[{{N}_{2}}=0.2N\] volume of solution after addition of water, \[{{V}_{2}}=10+x\] Normality of solution, \[{{N}_{2}}=0.001\text{ }M=0.002\text{ }N\] (\[\because \] for\[N{{a}_{2}}C{{O}_{3}},N=2M\]) \[\Rightarrow \] \[{{N}_{1}}{{V}_{1}}={{N}_{2}}{{V}_{2}}\] \[10\times 0.2=(10+x)\times 0.002\] \[x=990\,mL\]
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