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Chapter 8: Problem 135

The concentration of same aqueous solution of glucose is determined by two students-Sawan and Gautam. Sawan reported the concentration as \(20 \%\) (w/w) and Gautam reported the concentration as \(25 \%(\mathrm{w} / \mathrm{v}) .\) If both the concentrations are correct, then the density of solution is (a) \(0.8 \mathrm{~g} / \mathrm{ml}\) (b) \(1.0 \mathrm{~g} / \mathrm{ml}\) (c) \(1.25 \mathrm{~g} / \mathrm{ml}\) (d) \(1.33 \mathrm{~g} / \mathrm{m} 1\)

Short Answer

Expert verified
a) 0.8 g/ml b) 1.0 g/ml c) 1.2 g/ml d) 1.5 g/ml Answer: b) 1.0 g/ml

Step by step solution

01

Analyze the \(20 \%\) (w/w) concentration

Since the concentration is reported as \(20 \%\) (w/w), this means that \(20 g\) of glucose is present in \(100 g\) of the solution. To find the density of the solution, we need to determine the mass and volume of the solution.
02

Analyze the \(25 \%(\mathrm{w} / \mathrm{v})\) concentration

Since the concentration is reported as \(25 \%(\mathrm{w} / \mathrm{v})\), this means that \(25 g\) of glucose is present in \(100 ml\) of the solution. To find the density of the solution, we need the mass of the solution present in \(100 ml\).
03

Calculate the mass of the solution from the \(20 \%\) (w/w) concentration

From the given concentration (\(20 \%\) (w/w)), we know that there are \(20 g\) of glucose in \(100 g\) of solution. So, the mass of water in the solution will be \(100 g - 20 g = 80 g\). Thus, the mass of the solution is \(100 g\).
04

Calculate the volume of the solution from the \(25 \%(\mathrm{w} / \mathrm{v})\) concentration

From the given concentration (\(25 \%(\mathrm{w} / \mathrm{v})\)), we know there are \(25 g\) of glucose in \(100 ml\) of the solution.
05

Find the density of the solution

Since density is defined as the mass of the solution divided by its volume, we can now use the information from Steps 3 and 4 to calculate the density of the solution: Density = \(\frac{100 g}{100 ml}\) = \(1.0 \mathrm{~g} / \mathrm{ml}\) Therefore, the correct answer is (b) \(1.0 \mathrm{~g} / \mathrm{ml}\).

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Most popular questions from this chapter

It was found from the chemical analysis of a gas that it has two hydrogen atoms for each carbon atom. At \(0^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\), its density is \(1.25 \mathrm{~g}\) per litre. The formula of the gas would be (a) \(\mathrm{CH}_{2}\) (b) \(\mathrm{C}_{2} \mathrm{H}_{4}\) (c) \(\mathrm{C}_{2} \mathrm{H}_{6}\) (d) \(\mathrm{C}_{4} \mathrm{H}_{8}\)

When \(0.031\) of a mixture of hydrogen and oxygen was exploded, \(0.0031\) of oxygen remained. The initial mixture contains (by volume) (a) \(60 \% \mathrm{O}_{2}\) (b) \(40 \% \mathrm{O}_{2}\) (c) \(50 \% \mathrm{O}_{2}\) (d) \(30 \% \mathrm{O}_{2}\)

A quantity of \(0.25 \mathrm{~g}\) of a substance when vaporized displaced \(50 \mathrm{~cm}^{3}\) of air at \(0^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm} .\) The gram molecular mass of the substance will be (a) \(50 \mathrm{~g}\) (b) \(100 \mathrm{~g}\) (c) \(112 \mathrm{~g}\) (d) \(127.5 \mathrm{~g}\)

What volume of \(0.8 \mathrm{M}-\mathrm{AlCl}_{3}\) solution should be mixed with \(50 \mathrm{ml}\) of \(0.2 \mathrm{M}-\mathrm{CaCl}_{2}\) solution to get a solution of chloride ion concentration equal to \(0.6 \mathrm{M} ?\) (a) \(5.56 \mathrm{ml}\) (c) \(50 \mathrm{ml}\)

A quantity of \(5 \mathrm{~g}\) of a crystalline salt when rendered anhydrous lost \(1.8 \mathrm{~g}\) of water. The formula mass of the anhydrous salt is 160 . The number of molecules of water of crystallization in the salt is (a) 3 (b) 5 (c) 2 (d) 1
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