A solution of glucose in water is labelled as 10% w/w, what would be the molality and mole fraction of each component in the solution? If the density of solution is 1.2 g mL-1, then what shall be the molarity of the solution?
10% w/w solution of glucose in water means that 10 g of glucose in present in 100 g of the solution i.e., 10 g of glucose is present in (100 - 10) g = 90 g of water.
Molar mass of glucose (C6H12O6) = 6 × 12 + 12 × 1 + 6 × 16 = 180 g mol - 1
Then, number of moles of glucose = 10 / 180 mol
= 0.056 mol
∴ Molality of solution = 0.056 mol / 0.09kg = 0.62 m
Number of moles of water = 90g / 18g mol-1 = 5 mol
Mole fraction of glucose (xg) = 0.056 / ( 0.056+5) = 0.011
And, mole fraction of water xw = 1 - xg
= 1 - 0.011 = 0.989
If the density of the solution is 1.2 g mL - 1, then the volume of the 100 g solution can be given as:
= 100g / 1.2g mL-1
= 83.33 mL
=83.33 x 10-3 L
∴ Molarity of the solution = 0.056 mol / 83.33 x 10-3 L
= 0.67 M
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(v) Dichloridobis(ethane-1,2-diamine)platinum(IV) nitrate
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(iii) 2-Hydroxycyclopentane carbaldehyde
(iv) 4-Oxopentanal
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(ii) Ethanol to but-1-yne
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(v) Benzene to 4-bromonitrobenzene
(vi) Benzyl alcohol to 2-phenylethanoic acid
(vii) Ethanol to propanenitrile
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(x) 2-Methyl-1-propene to 2-chloro-2-methylpropane
(xi) Ethyl chloride to propanoic acid
(xii) But-1-ene to n-butyliodide
(xiii) 2-Chloropropane to 1-propanol
(xiv) Isopropyl alcohol to iodoform
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(xvii) Chloroethane to butane
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Thx for helping me ð
1000/90 can be written as 0.09
1000/90 can be written as 0.09
Why 19 was divided by 18.instead of 180
From where 0.9 is there
0.9 kha s aaya h
When we get 0.01
Hey
What if density is not given
Very interesting question