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.
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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
