JEE Mains · Chemistry · STD 11 - 1. Some basic concept of chemistry
A solution of two components containing \(n_{1}\) moles of the \(1^{\text {st }}\) component and \(n_{2}\) moles of the \(2^{\text {nd }}\) component is prepared. \(M _{1}\) and \(M _{2}\) are the molecular weights of component \(1\) and \(2\) respectively. If \(d\) is the density of the solution in \(g\, mL ^{-1}, C _{2}\) is the molarity and \(x _{2}\) is the mole fraction of the \(2^{\text {nd }}\) component, then \(C_{2}\) can be expressed as
- A \(C _{2}=\frac{1000 x _{2}}{ M _{1}+ x _{2}\left( M _{2}- M _{1}\right)}\)
- B \(C _{2}=\frac{ d x _{2}}{ M _{2}+ x _{2}\left( M _{2}- M _{1}\right)}\)
- C \(C _{2}=\frac{ d x _{1}}{ M _{2}+ x _{2}\left( M _{2}- M _{1}\right)}\)
- D \(C _{2}=\frac{1000 dx _{2}}{ M _{1}+ x _{2}\left( M _{2}- M _{1}\right)}\)
Answer & Solution
Correct Answer
(D) \(C _{2}=\frac{1000 dx _{2}}{ M _{1}+ x _{2}\left( M _{2}- M _{1}\right)}\)
Step-by-step Solution
Detailed explanation
\(C_{2}=\frac{x_{2}}{\left[x_{2} M_{2}+\left(1-x_{2}\right) M_{2}\right] / C l} \times 1000\) \(C_{2}=\frac{1000 d x_{2}}{M_{1}+\left(M_{2}-M_{1}\right) x_{2}}\)
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