Calculate
the molarity of each of the following solutions: (a) 30 g of Co(NO3)2.6H2O in 4.3 L of solution (b) 30 mL of 0.5 M H2SO4 diluted to 500 mL.
Solution
(a)
Given that
mass of Co (NO3)2.6H2O = 30 g
Volume of solution = 4.3 liter
Molar mass of Co(NO3)2.6H2O
= 59 + 2(14 + 3 × 16) + 6 (16 +1 × 2) = 291 g mol−1
Use the formula
Moles of Co(NO3)2.6H2O
= 30/291 = 0.103 mol
Use the formula
Molarity of Co(NO3)2.6H2O
= 0.103/4.3= 0.023 M
(b)
Given that
Volume of
H2SO4 (V1) = 30 ml)
Diluted
volume of H2SO4 (V2) = 500 ml
Number of moles present in 1000 mL
of 0.5 M H2SO4 = 0.5 mol
Initial molarity M1= 0.5
M
We have to find M2
Use the formula
M2V2 = M1V1
plug the values we get
M2 × 500
= 0.5 × 30
M2 = 15/ 500 = 0.03 M
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ReplyDeleteGreat explanation of the molarity calculations your step by step approach makes it much easier to understand how to convert mass to moles and apply the dilution formula correctly for both parts. The breakdown of Co(NO₃)₂·6H₂O and the use of M₁V₁ = M₂V₂ were especially clear and well presented also, I really appreciate how smoothly you highlighted allpanel in your discussion it fits naturally with the topic and adds a nice, professional touch to your overall presentation.
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