Question

Number of HCl molecules present in 10 ml of 0.1 M solution?

Answer 1

Molarity = moles of solute/liters of solution

or, for our specific purpose here

moles of solute (HCl) = liters of solution * Molarity

10 ml = 0.01 liters

moles of solute (HCl) = (0.01 liters)*(0.1 M HCl)

= 0.001 moles of HCl

Answer 2

Answer:

The number of HCl molecules in 10ml 0f 0.1M solution is 6.023 ×10²⁰ molecules.

Step by step explanation:

Given that,

      Molarity of solution = 0.1M,

                   $Molarity=\frac{n_{x} }{V}$

    Where nₓ = moles of solute of HCl

                V= volume of the solution $= 10ml= 0.01L$

                    $0.1M=\frac{n_{x}}{0.01L}}\\ \\ (0.1M)(0.01L)=n_{x}$

      ∴      number of moles of solute(HCl) = 0.001

  No. of molecules = (moles).(Avogadro no.)

                               = 0.001×6.023×10²³

                                = 6.023×10²⁰ molecules

    Therefore, number of HCl molecules present in 10ml of 0.1M solution is 6.023×10²⁰ molecules.