Answer to Question #32318 in Inorganic Chemistry for maharaj

Question #32318

how much NaNO3 must be weighted out to make 50ml of aqueous sol. containing 70mg per
ml?

Expert's answer

The concentration of Na is 70 mg / mL and you have 50 mL, soyou have 70 mg/mL * 50 mL = 3500 mg.
The atomic mass of Na is 23 g/mole.Calculate how many moles of Na you have in 3500 mg (3.g): 3,5/23 = 0.15 moles.
NaNO₃ dissociates into Na⁺ and NO₃^- so the moles of NaNO₃ is the same as molesof Na⁺. To get the mass of NaNO₃, we can multiply the moles of NaNO₃ by the molecular weight of NaNO₃: 0,15*85 = 12,75g.

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Assignment Expert

31.08.15, 10:33

Dear [b]Washim Chowdhury, [/b]this answer is also right. At first step amount of Na was found (0.15 moles). The amount of sodium equals the amount of sodium nitrate. So, to find the mass of sodium nitrate you can multiply the moles (0.15 moles) by the molecular weight of sodium nitrate. You'll get the mass of 0.15 moles of sodium nitrate that equals 0.15 moles (or 3.5 g in 50 ml) of sodium ions. Both methods of calculation give the same result.

Washim Chowdhury

30.08.15, 20:33

As you say in last sentence, multiply the moles of NaNO3, but you did with moles of Na you got earlier. I know, you have followed the answer from yahoo. Interested ones, may visit for this answer - https://www.meritnation.com/ask-answer/question/how-much-nano3-must-be-weighed-out-to-make-50-ml-of-an-aqueo/chemistry/5488184

Answer to Question #32318 in Inorganic Chemistry for maharaj

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