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Furkat [3]
3 years ago
13

Calculate the molarity when 1.24 g AgNO3 is dissolved to make 125 mL solution. Round to two significant digits.

Chemistry
2 answers:
Goshia [24]3 years ago
4 0

Answer:

Molarity=0.0058M

Explanation:

Given the weight of AgNO_{3} is(w) 1.24g

We know that the molecular weight of AgNO_{3} is(W) 169.87g/mol.

Given the volume of the solution is(V) 125ml.

Here solute is silver nitrate and the solvent is water.

We know that

Molarity=\frac{number of moles of solute}{total volume of the solution int litres}

We know that number of moles=weight/molecular weight

n= \frac{1.24}{169.87}=   7.3*10^{-3}

volume of the solution(V)=125/1000=0.125L

Now

Molarity=n/V

M=  \frac{7.3*10^{-3} }{0.125}   =0.00584M

kicyunya [14]3 years ago
3 0
<h2>0.058 mol/L</h2>

Explanation:

       \textrm{Molarity of a solution}=\frac{\textrm{Number of moles of solvent}}{\textrm{Volume of solution in Liters}}

       We are given a 125mL solution which contains 1.24g of AgNO_{3}.

       To calculate number of moles of AgNO_{3} present, we need to find it's molar mass from charts. Molar Mass of AgNO_{3} is known to be 169.87\frac{g}{Mol}.

       Number of moles of AgNO_{3} present = \frac{\textrm{Given weight}}{\textrm{Molar Mass}}\textrm{ = }\frac{1.24g}{169.87\frac{g}{Mol}}\textrm{ = }0.0073mol

       Molarity = \frac{0.0073mol}{0.125L}=0.0584\frac{mol}{L}

∴ Molarity of given AgNO_{3} solution = 0.058\frac{mol}{L}

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igor_vitrenko [27]

Answer:

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Explanation:

The following data were obtained from the question:

Length of whale (in mm) = 28578 mm

Length of whale (in km) =?

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1 mm = 1×10¯⁶ Km

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