Question

3.01×10²²molecules of glucose (C6 H12 O6) are present in 100ml of water ( density of solution = 1 g/ml).the molality of the solution is
a) 0.25m
b) 1m
c) 1.25 m
d) 0.5m​

Answer 1

Answer :-

Molality of the solution is 0.5 m [Option.d]

Explanation :-

Firstly, let's calculate the number of moles of solute i.e. glucose present. This can be calculated as :-

$\boxed{\bf{\textbf{Moles} = \dfrac{Number \: of \: molecules}{Avogadro \: Number}}}$

$\tt\implies Moles = \dfrac{3.01 \times 10^{22}}{6.02 \times 10^{23}}\\\\\tt\implies Moles = \dfrac{1}{20}\\\\\tt\implies Moles = 0.05$

The next step is, we have to calculate the mass of the solvent. Mass of solvent is :-

$\tt\implies Mass = Volume \times Density\\\\\tt\implies Mass = 100 \times 1\\\\\tt\implies Mass = 100 \: g\\\\\tt\implies Mass = 0.1 \: kg$

Finally, let's put the values to obtain molality of the solution .

$\boxed{\bf{\textbf{Molality} = \dfrac{Moles \: of \: solute}{Mass \: of \: solvent \: in \: kg}}}$

$\tt\implies Molality = \dfrac{0.05}{0.1}\\\\\tt\implies\underline{Molality = 0.5 \: m}$

Answer 2

Answer:

Given :-

• 3.01 × 10²² molecules of glucose are present in 100 ml of water (density of solution is 1 g/ml).

To Find :-

• What is the molality of the solution.

Solution :-

First, we have to find the numbers of moles :

As we know that :

$\clubsuit$ Number of Moles Formula :

$\footnotesize \bigstar \: \: \sf\boxed{\bold{\pink{Number\: of\: Moles =\: \dfrac{Number\: of\: Molecules}{Avogadro's\: Number}}}}\: \: \: \bigstar$

Given :

• Avogadro's Number = 6.02 × 10²³
• Number of molecules = 3.01 × 10²²

According to the question by using the formula we get,

$\implies \sf Number\: of\: Moles =\: \dfrac{3.01 \times 10^{22}}{6.02 \times 10^{23}}$

$\implies \sf Number\: of\: Moles =\: \dfrac{0.5}{10}$

$\implies \sf\bold{\blue{Number\: of\: Moles =\: 0.05}}$

Now, we have to find the mass of the solution :

As we know that :

$\footnotesize \bigstar \: \: \sf\boxed{\bold{\pink{Density\: of\: Solvent =\: \dfrac{Mass\: of\: Solvent}{Volume\: of\: Solvent}}}}\: \: \: \bigstar$

Given :

• Density = 1 g/ml
• Volume = 100 ml

According to the question by using the formula we get,

$\implies \sf 1 =\: \dfrac{Mass\: of\: Solvent}{100}$

By doing cross multiplication we get,

$\implies \sf Mass\: of\: Solvent =\ 100 \times 1$

$\implies \sf\bold{\green{Mass\: of\: Solvent =\: 100\: g}}$

Now, we have to convert the mass of solution g into kg :

$\implies \bf Mass\: of\: Solvent =\: 100\: g$

$\implies \sf Mass\: of\: Solvent =\: \dfrac{100}{1000}\: kg$

$\implies \sf\bold{\purple{Mass\: of\: Solvent =\: 0.1\: kg}}$

Now, we have to find the molality of the solution :

As we know that :

$\footnotesize \bigstar \: \: \sf\boxed{\bold{\pink{Molality\: (m) =\: \dfrac{Moles\: of\: Solute}{Mass\: of\: Solvent\: (Kilogram)}}}}\: \: \: \bigstar$

Given :

• Moles of Solute = 0.05
• Mass of Solvent = 0.1 kg

According to the question by using the formula we get,

$\dashrightarrow \sf Molality\: (m) =\: \dfrac{0.05}{0.1}$

$\dashrightarrow \sf\bold{\red{Molality\: (m) =\: 0.5\: m}}$

$\therefore$ The molality of the solution is 0.5 m .

Hence, the correct options is option no (d) 0.5 m .