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November 01, 2025, 12:59:13 pm
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Author Topic: A123 Chem Question.  (Read 1366 times)  Share 

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abd123

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A123 Chem Question.
« on: March 29, 2011, 08:34:21 pm »
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Q11. To prevent a gum disease called scurvy, the minimum daily vitamin
C (C6 H8 06) required is 60mg.
(a) How many moles of vitamin C is this?
(b) How many molecules is this?
(c) If 10g of spinach is found to contain 1.2 x 10^-5g
of Vitamin C, how much spinach must be eaten to attain the
minimum daily requirement?
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Souljette_93

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Re: A123 Chem Question.
« Reply #1 on: March 29, 2011, 08:56:27 pm »
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First q: convert mg to grams, so that would be 60 *10^-3 g

Find mole, n=m/mr, equal to 0.0003409

Second qs= multiply by avagadros no


Third qs= use ratios, so if 10 g gives me 1.2*10^-5
            How much will.     X g.               60*10^-3

X=0.000005
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jasoN-

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Re: A123 Chem Question.
« Reply #2 on: March 29, 2011, 09:34:56 pm »
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ZnSO4*XH2O ---> ZnSO4 + XH2O
   m=8.2g             Mr=161
                          m=4.6g
n(ZnSO4) yielded = m/Mr=0.0285714mol

n(ZnSO4*XH2O) = n(ZNSO4) = 0.0285714

Mr(ZnSO4*XH2O) = m/n = 8.2/0.0285714 = 287g/mol

We know that Mr(ZnSO4) = 161
So: Mr(XH2O) = 287 - 161 = 126
Mr(H2O) = 18
So X = 126/18 = 7
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Re: A123 Chem Question.
« Reply #3 on: June 19, 2011, 02:41:19 pm »
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m(MgSO4)=0.461g
m(H2O)=0.942-0.461= 0.481g
Then use mole ratios so: n(MgSO4):n(H2O)
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