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Question: Commercially available \(conc.HCl\) contains 38% \(HCl\) by mass, what is the molarity of the soluti...

Commercially available conc.HClconc.HCl contains 38% HClHCl by mass, what is the molarity of the solution, density of solution=1.19g ml−11.19g\text{ }m{{l}^{-1}}. What volume of conc. HCl is required to make 10 L of a 0.10M HCl?

Explanation

Solution

First we have to find out the mass of solution by using the density formula as; Density=MassVolumeDensity=\dfrac{Mass}{Volume} and then we can easily find the molarity of the solution by; molarity=molarity = no of moles of the solutevolume of the solution in 1 liter\dfrac{no\text{ }of\text{ }moles\text{ }of\text{ }the\text{ }solute}{volume\text{ }of\text{ }the\text{ }solution\text{ }in\text{ }1\text{ }liter} and hence, by using this molarity we can find the volume of conc. HCl is required to make 10 L of a 0.10M HCl by applying the molarity equation. Now solve it.

Complete step by step solution:
First of all let’s discuss what is molarity. By the term molarity, we mean the number of moles of the solute to the total volume of the solution in one litre or thousand milliliter of the solution.
i.e.
molarity=molarity = no of moles of the solutevolume of the solution in 1 liter\dfrac{no\text{ }of\text{ }moles\text{ }of\text{ }the\text{ }solute}{volume\text{ }of\text{ }the\text{ }solution\text{ }in\text{ }1\text{ }liter}
Now considering the statement;
We can calculate the molarity of the solution by using the formula as;
molarity=molarity = no of moles of the solutevolume of the solution in 1 liter\dfrac{no\text{ }of\text{ }moles\text{ }of\text{ }the\text{ }solute}{volume\text{ }of\text{ }the\text{ }solution\text{ }in\text{ }1\text{ }liter}-----------(1)
38% HCl by mass means 38g of HCl in 100g of the solution.
density of solution=1.19g ml−11.19g\text{ }m{{l}^{-1}}(given)
mass of the solution =100g100 g
As we know that:
Density=MassVolume  1.19=100volume volume=1001.19  = 84 ml  = 841000L(1 l=1000 ml) \begin{aligned} & Density=\dfrac{Mass}{Volume} \\\ & \text{ 1}\text{.19}=\dfrac{100}{volume} \\\ & volume=\dfrac{100}{1.19} \\\ & \text{ = 84 }ml \\\ & \text{ = }\dfrac{84}{1000}L(1\text{ }l=1000\text{ }ml) \\\ \end{aligned}
Number of moles of HCl=given massmolar mass\dfrac{given\text{ }mass}{molar\text{ }mass}
Given mass= 38g
Molar mass of HCl= 36.5g
Thus,
Number of moles of HCl=3836.5=1.04 mole\dfrac{38}{36.5}=1.04\text{ }mole
Now, put all these values in equation (1), we get;
molarity=molarity = 1.04×100084=12.4 mol l−1\dfrac{1.04\times 1000}{84}=12.4\text{ }mol\text{ }{{l}^{-1}}
Therefore, the molarity of the HClHCl solution which contains 38% HClHCl by mass and having the density of solution as 1.19g ml−11.19g\text{ }m{{l}^{-1}}is =12.4 mol l−1=12.4\text{ }mol\text{ }{{l}^{-1}}.
Now, coming to the next part, we have been the :
Molarity of HCl sampleM1{{M}_{1}} =12.4 mol l−1=12.4\text{ }mol\text{ }{{l}^{-1}}
Volume of HCl of sample =V1{{V}_{1}}
Molarity of sample which is to preparedM2{{M}_{2}} =0.10 mol l−1=0.10\text{ }mol\text{ }{{l}^{-1}}
Volume of HCl which is to prepared V2{{V}_{2}} =1 L=1000 ml=1\text{ L=1000 }ml
Therefore, now using the molarity equation as;
M1V1=M2V2  V1=0.10×100012.38  = 8.07 ml \begin{aligned} & {{M}_{1}}{{V}_{1}}={{M}_{2}}{{V}_{2}} \\\ & \text{ }{{V}_{1}}=\dfrac{0.10\times 1000}{12.38} \\\ & \text{ = 8}\text{.07 }ml \\\ \end{aligned}

Therefore, volume of conc. HCl is required to make 10 L of a 0.10M HCl is  = 8.07 ml\text{ = 8}\text{.07 }ml.

Note: Molarity of the solution changes with the change in the temperature because volume changes with the temperature and hence, molarity is also changing while on the other hand, molality which may be defined as the number of moles of the solute to the total mass of the solvent in 1 kg; has no effect of temperature on it.