Question

Solutions containing $23gHCOOH$ is/are:
This question has multiple correct options.
A.\text{46g of 70%} \left( {w/v} \right){\text{ }}HCOOH\left( {{d_{solution}} = 1.40g/mL} \right)
B.$\text{50g of 10MHCOOH}\left( {{d_{solution}} = 1g/mL} \right)$
C.\text{50g of 25%} \left( {w/w} \right)HCOOH
D.$\text{46g of 5MHCOOH}\left( {{d_{solution}} = 1g/mL} \right)$

Answer 1

We need to evaluate each of the given statements which correspond to a given amount of the compound. Accurate stoichiometric calculations are to be done. While making a solution, there are different ways to express the concentration of a solution. The most common ones are Molarity (M), Weight by volume (w/v), weight by weight (w/w) and volume by volume (v/v).

Complete step by step answer:
We need to know that the different ways to express the concentration of a solution are discussed below:
Molarity (M): Molarity is defined as the number of moles of solute in one liter of solution.
Weight by volume (w/v): Weight by volume concentration is expressed as weight of the solute in grams in one milliliter of solvent. w/v is often used when a dry solute is weighed out and added to a liquid solvent.
Weight by weight (w/w): Weight by weight is expressed as weight of solute in grams in hundred grams of the solution. It is often used when a dry solute is mixed in a dry solvent.
Volume by volume (v/v): Volume by volume is expressed as volume of solute in litres present in one litre of the solution.
Now we evaluate each of the given stoichiometric statements of the concentrations.
Given that Solutions contain $23g$ of $HCOOH$. The molecular weight of $HCOOH$ is calculated to be $46g/mol$. We are to analyse each of the given statements and conclude if it corresponds to $23g$ of $HCOOH$ or not.
\text{46g of 70%} \left( {w/v} \right){\text{ }}HCOOH\left( {{d_{solution}} = 1.40g/mL} \right)
\text{46g of 70%} \left( {w/v} \right)HCOOH = \dfrac{{46 \times 70}}{{1.40 \times 100}} = 23g
Therefore, \text{46g of 70%} \left( {w/v} \right){\text{ }}HCOOH with $\left( {{d_{solution}} = 1.40g/mL} \right)$ correspond to $23g$ of $HCOOH$. Hence, this is a correct option.
$\text{50g of 10MHCOOH}\left( {{d_{solution}} = 1g/mL} \right)$
Since the density of the solution is $1g/ml$ , therefore $50g$ of solution will correspond to $50ml$ of the solution. We now calculate the number of moles in $10M$ solution.
$50ml$ of 10M $HCOOH = \dfrac{{50}}{{1000}} \times 10 = 0.5moles$
In $46g/mol$ of $HCOOH$ , weight of $HCOOH = 46 \times 0.5 = 2.3g$.
Therefore, $50g$ of $10MHCOOH$ with density, ${d_{solution}} = 1g/mol$ corresponds to $23g$ of $HCOOH$ . Hence, this is a correct option.
\text{50g of 25%} \left( {w/w} \right)HCOOH
\text{50g of 25%} \left( {w/w} \right)HCOOH = \dfrac{{50 \times 25}}{{100}} = 12.5g
Since \text{50g of 25%} \left( {w/w} \right)HCOOH does not correspond to $23g$ of $HCOOH$, therefore it is an incorrect option.
$46g$ of $5HCOOH({d_{solution}} = 1g/mol)$: Since the density is $1g/mol$ , therefore $46g$ of solution will correspond to $46mL$ of the solution.
We now calculate the number of moles in $5M$ solution.$46ml$ of 10M $HCOOH = \dfrac{{46}}{{1000}} \times 5 = 0.23moles$.
In $46ml$ of 10M $HCOOH$, weight of $HCOOH = 46 \times 0.23 = 10.6g$.
Therefore, $46ml$ of 10M $HCOOH$ with density, ${d_{solution}} = 1g/ml$ does not correspond to $23g$ of $HCOOH$ . Hence, this is an incorrect option.
Therefore, the correct options are options (A) and (B).

Note:
We must be noted that different units are used to express the concentrations of a solution depending on the application and the most important parameter to describe the concentration of a solution is to know the molecular weight of the solute. Molarity is most widely used to describe concentrations of solutions. Another way of expressing concentration of solutions involving gases is mole fraction. Molality is also used to describe moles of solute per kg of the solvent.