Question

The amount of sugar \[{C_{12}}{H_{22}}{O_{11}}\]required to prepare \[{\text{2L}}\] for its \[{\text{0}}{\text{.1M}}\]
A.${\text{68}}{\text{.4g}}$
B.${\text{17}}{\text{.1g}}$
C.${\text{34}}{\text{.2g}}$
D.${\text{136}}{\text{.8g}}$

Answer 1

Hint: To answer this question, you should recall the concept of concentration of a solution. Find the number of moles using the molarity given which will give you the value of the weight of sugar.
The formula used:
\[{\text{Molarity = }}\dfrac{{{{{\text{(n)}}}_{{\text{solute}}}}}}{{{{\text{V}}_{{\text{solute}}}}{\text{(in litre)}}}}\]where \[{\text{n}}\]is the number of moles and \[{\text{V}}\]is the volume of solvent.

Complete step by step answer:
Molarity is a commonly used method to express the concentration of molarity. It is the number of moles of solute dissolved in one litre of a solution.
Substituting the value in the formula of molarity we get,
\[0.1 = \dfrac{{{\text{Amount of solute}}}}{{342 \times 2}}\].
Solving and rearranging for solute i.e. sugar
$\therefore $Amount of sugar \[\left( {{{\text{C}}_{{\text{12}}}}{{\text{H}}_{{\text{22}}}}{{\text{O}}_{{\text{11}}}}} \right) = \]\[68.4{\text{gram}}\].

Hence, option A is correct.

Additional information:
Other concentration terms used are:
Concentration in Parts Per Million (ppm) The parts of a component per million parts (\[{10^6}\]) of the solution.
\[{\text{ppm(A) = }}\dfrac{{{\text{Mass of A}}}}{{{\text{Total mass of the solution}}}} \times {10^6}\]
$\% {\text{w/w}}$ is weight concentration of a solution: If a solution is labeled as \[10\% \]glucose in water by mass, it refers to that 10g of glucose is dissolved in 90 g of water resulting in 100g of solution. $\% {\text{v/v}}$ is the volume concentration of a solution: it refers that if 50 mL of acetic acid is added to 50 mL of water, the acetic acid is labelled as \[50\% {\text{v/v}}\]. $\% {\text{w/v}}$ is the mass concentration of a solution: if x grams/ml of solute are present in solution it means x gram of solute X is dissolved in 100ml of solution. Suppose a solution contains solute A and solvent B, then its mass percentage is expressed as:
\[{\text{Mass % of A = }}\dfrac{{{\text{Mass of component A in the solution}}}}{{{\text{Total mass of the Solution}}}}{{ \times 100}}\]
and Volume Percentage (V/V) can be written as: \[{\text{volume% of A = }}\dfrac{{{\text{Volume of component A in the solution}}}}{{{\text{Total volume of the solution}}}}{{ \times 100}}\]
Molality (m): Molality establishes a relationship between moles of solute and the mass of solvent. It is given by moles of solute dissolved per kg of the solvent. The molality formula is as given- \[{\text{Molality(m) = }}\dfrac{{{\text{Moles of solute}}}}{{{\text{Mass of solvent in kg}}}}\]

Note:
Normality: It is defined as the number of gram equivalents of solute present in one litre of the solution.
Mole Fraction: It gives a unitless value and is defined as the ratio of moles of one component to the total moles present in the solution. Mole fraction = $\dfrac{{{X_{\text{A}}}}}{{{X_{\text{A}}} + {X_B}}}$(from the above definition) where ${X_{\text{A}}}$is no. of moles of glucose and \[{X_{\text{B}}}\]is the no. of moles of solvent