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1 year ago

Suppose 0.258g of lead(II) acetate is dissolved in 250.mL of a 13.0 m M aqueous solution of ammonium sulfate.

1 answer:

8 0

Molarity is the estimation of the solute concentration in the solution. The final molar concentration of the acetate anion in the solution is 0.006 M.

<h3>What is Molarity?</h3>

Molarity is the ratio of the moles to the volume in liters.

Given,

Mass of lead (II) acetate = 0.258 gm

Volume of solution = 0.25 L

Moles are calculated as:

Moles = mass ÷ molar mass

= 0.258 g ÷ 325.29 g/mol

= 0.00079

Molarity is calculated as:

M = moles ÷ volume

= 0.00079 ÷ 0.25

= 0.0032

The molar concentration of the acetate ions: 2 × 0.0032 = 0.0064 M

Therefore, the molarity of the acetate anion in the solution is 0.006 M.

Learn more about molarity here:

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$\bold{\huge{\orange{\underline{ Solution}}}}$

$\bold{\underline{ Given :- }}$

We have 250g of liquid water and it needs to be cool at temperature from 100° C to 0° C
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$\bold{\underline{ To \: Find :- }}$

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$\bold{\underline{ Let's \:Begin:- }}$

We know that,

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$\sf{\red{ Q = mcΔT }}$

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$\sf{ Q = 250 × 4.180 × - 100 }$

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$\bold{\underline{ Now :- }}$

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