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A \(40 \%\) by mass sucrose solution is heated till, it becomes \(50 \%\) by mass. Calculate the mass of water lost from \(100 \mathrm{~g}\) of the solution is
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The correct answer is:
\(20 \mathrm{~g}\)
\(40 \%\) sucrose solution means it contains \(60 \%\) water. After heating, till \(50 \%\) by mass sucrose remains.
Thus, \(\%\) water lost \(=\frac{0.4}{0.5} \times 100=80\)
Water lost \(=100-80=20 \mathrm{~g}\)
Hence, statement (c) is correct.
Thus, \(\%\) water lost \(=\frac{0.4}{0.5} \times 100=80\)
Water lost \(=100-80=20 \mathrm{~g}\)
Hence, statement (c) is correct.
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