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------------------------------------------------ How much NaNO₃ is needed to prepare 225 mL of a 1.55 M solution of NaNO₃?

Answer :

29.64 gram of NaNO₃ is needed to prepare 225 ml of a 1.55 M solution of NaNO₃.

How we calculate molarity?

Molarity of given NaNO₃ solution can be calculated as M =n/V, where

M = molarity = 1.55M (given)

n = no. of moles = to find?

V = volume = 225mL = 0.225L

From the above formula we can calculate moles of NaNO₃ as follow:

n = M × V

n = 1.55 × 0.225 = 0.34875 moles

Now can calculate the required mass of NaNO₃ by using below formula:

n =W/M, where

W = required mass = ?

M =molar mass = 84.99 g/mol

Putting values in above equation, we get

W = n × M

W = 0.348 mole × 84.99 g/mol = 29.64 g

Hence, 29.64 gram of NaNO₃ is needed.

To learn more about molarity, visit below link:

https://brainly.com/question/10608366

Answer:

= 29.64 g NaNO3

Explanation:

Molarity is given by the formula;

Molarity = Moles/Volume in liters

Therefore;

Number of moles = Molarity × Volume in liters

= 1.55 M × 0.225 L

= 0.34875 moles NaNO3

Thus; 0.34875 moles of NaNO3 is needed equivalent to;

= 0.34875 moles × 84.99 g/mol

= 29.64 g