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GUJCET · Chemistry · Solutions
Calculate the mass of Glucose \(\left( C _6 H _{12} O _6\right)\) required in making 2.5 kg of 0.25 molal aqueous solution.[Atomic wt : \(H =1, O =16, C =12 amu\) ]
- A 135.0 g
- B 107.65 g
- C 90.0 g
- D 112.5 g
Answer & Solution
Correct Answer
(B) 107.65 g
Step-by-step Solution
Detailed explanation
\(M_{C_6H_{12}O_6} = 6(12) + 12(1) + 6(16) = 180 \text{ g/mol}\) \(m = \frac{\text{moles of solute}}{\text{mass of solvent (kg)}}\)
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