解:(1)由過氧乙酸的化學式為CH
3COOOH可知碳、氫、氧原子個數(shù)比為:2:4:3;
(2)過氧乙酸中碳、氫、氧元素的質(zhì)量比為:(12×2):4:(16×3)=6:1:12;
(3)設應取用此瓶過氧乙酸的質(zhì)量為x,則
x×40%=20g×10%
解得,x=5g;
(4)將500g該溶液與100g10%的過氧乙酸溶液混合,所得溶液的溶質(zhì)質(zhì)量分數(shù)=
×100%=35%.
故答案為:
(1)2:4:3;
(2)6:1:12;
(3)5g;
(4)35%.
分析:(1)有過氧乙酸的化學式得出過氧乙酸中原子個數(shù)比;
(2)物質(zhì)中各元素的質(zhì)量比為各元素原子的相對原子質(zhì)量與原子個數(shù)的乘積之比;
(3)配制過程中溶質(zhì)質(zhì)量不變;由稀溶液的溶液質(zhì)量和溶質(zhì)質(zhì)量分數(shù)可以計算出溶質(zhì)質(zhì)量,由溶質(zhì)質(zhì)量和濃溶液的溶質(zhì)質(zhì)量分數(shù)可以計算出濃溶液的質(zhì)量;
(4)與100g10%的過氧乙酸溶液混合后,所得溶液的質(zhì)量為兩溶液的質(zhì)量和、溶液中溶質(zhì)的質(zhì)量為兩溶液中溶質(zhì)質(zhì)量之和,利用溶液中溶質(zhì)的質(zhì)量分數(shù)的計算公式,計算所得溶液的溶質(zhì)質(zhì)量分數(shù).
點評:本題難度不大,主要考查了有關化學式的計算及有關溶液中溶質(zhì)質(zhì)量的計算,培養(yǎng)學生的分析能力和解決問題的能力.