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3 years ago

How many moles of solute particles are present in 1 ml of aqueous 0.020 m (nh4)2co3?

1 answer:

6 0

Vs = 1.0 mL = 0.001 L
c((NH4)2CO3) = 0.02 M
n((NH4)2CO3) = ?

For the purpose, here we will use the next equation:

c=n/V ⇒ n=cxV

n((NH4)2CO3) = 0.02M x 0.001L

n((NH4)2CO3) = 2x10⁻⁵ mole of (NH4)2CO3 is presented in the solution

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Explanation:

We are given:

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To calculate hydroxide ion concentration, we use the equation to calculate pOH of the solution, which is:

$pOH=-\log[OH^-]$

We are given:

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Putting values in above equation, we get:

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To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of solution = $1.17\times 10^{-12}M$

Volume of solution = 1243 mL = 1.243 L (Conversion factor: 1 L = 1000 mL)

Putting values in above equation, we get:

$1.17\times 10^{-12}M=\frac{\text{Moles of }OH^-}{1.243L}\\\\\text{Moles of }OH^-=(1.17\times 10^{-12}mol/L\times 1.243L)=1.424\times 10^{-12}mol$

According to mole concept:

1 mole of a compound contains $6.022\times 10^{23}$ number of particles

So, $1.424\times 10^{-12}mol$ number of $OH^-$ will contain = $(1.424\times 10^{-12}\times 6.022\times 10^{23})=8.57\times 10^{11}$ number of ions

Hence, the number of $OH^-$ ions dissociated are $8.57\times 10^{11}$

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