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58问答库 > 实验室配制500ml1mol⼀L的稀硫酸溶液,需用密度为1.84g⼀cm3.质量分数为98%的浓硫酸

实验室配制500ml1mol⼀L的稀硫酸溶液,需用密度为1.84g⼀cm3.质量分数为98%的浓硫酸

2025-04-07 07:07:20

推荐回答（2个）

回答（1）：

需要浓硫酸的物质的量为n=0.5*1=0.5mol
质量为m=0.5mol*98g/mol=49g
设浓硫酸体积为V（mL）则其质量为m=p（密度）*v*98%
以上两式都是硫酸的质量，所以相等代入数值
49=1.84*98%*V
解得V=27.2ml

回答（2）：

设需要xml
1*0.5=(1000*1.84*98%)x/98,x=0.027L=27ml

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