Question

.将4.83 mol H2SO4溶于水配成1L密度为1.28g/cm3的硫酸溶液,求其物质的量浓度?硫酸的质量分数?

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Answer 1

【1】求其物质的量浓度? 解：C=n/V=4.83/1=4.83 (mol/L)
【2】硫酸的质量分数? 解：w = M*C/1000p = 98.08*4.83/1000*1.28 = 37%

Answer 2

1L=1000cm^3,所以硫酸溶液的质量为1.28g/cm^3x1x10^3cm^3=1280g。因为有4.83mol,所以H2SO4的质量为98g/molx4.83mol=473.54g,则质量分数为473.54/1280x100%=37% c=4.83mol/1L=4.83mol/L