How many grams of Fe(NO3)2 are required to make 400 mL of a solution with concentration 0.5 M Fe(NO3)2?
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Answer:
36 gm
Explanation:
Formula of Molarity=>
M=(w/m)×1000/V(in ml)
=>w=(M×m×V)/1000
Putting the values,
=>w=0.5×180×400/1000
=>w=36 gm (Ans)
sagarsadhna806:
thankyou
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