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how many grams of AgNO3 (MM=169.87) are needed to prepare 0.125 M solution in 250 mL of water?
a. 0.5 g
b. .03 g
c. 5.3 g
d. 84.9g​

Respuesta :

Ankit

Answer:

Correct answer-C 5.3g

Explanation:

Molarity = no. of moles of solute/ volume of of solution in litre

olarity = no. of moles of solute/ volume of of solution in litreno of moles of solute = 0.125×0.25=0.03125

olarity = no. of moles of solute/ volume of of solution in litreno of moles of solute = 0.125×0.25=0.03125one mole AgNO3 weighs 169.87 gm,

olarity = no. of moles of solute/ volume of of solution in litreno of moles of solute = 0.125×0.25=0.03125one mole AgNO3 weighs 169.87 gm,so the mass of 0.03125 moles of AgNO3 = 0.03125×169.87= 5.3084g

The grams of AgNO₃ are needed to prepare 0.125 M solution in 250 mL of water is 5.3 grams.

How do we calculate mass from moles?

Moles is a unit which is used to define any amount and it is calculated as:

n = W/M, where

W = required mass

M = molar mass

Given molarity of AgNO₃ = 0.125 M

Volume of water = 250mL = 0.25L

moles (n) from the molarity (M) will be calculated as:

n = M × V

n = (0.125)(0.25) = 0.0312 moles

Now we calculate the mass of AgNO₃ by using the above formula as:

W = (0.0312mol)(169.87g/mol)

W = 5.30 g

Hence option (c) is correct.

To know more about moles, visit the below link:

https://brainly.com/question/15374113

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