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<p class=WL1>개미산의 분자량은 46임.</p>
<p class=EL1>&nbsp;</p>
<p class=WL1>1. 20℃, 1기압 하에서 완전히 기화된 개미산 1㎖의 무게는?</p>
<p class=WL1>PV = nRT (∵n=몰수=무게/분자량)</p>
<p class=WL1>n = PV/(RT)</p>
<p class=WL1>무게/분자량 = PV / (RT)</p>
<p class=WL1>무게 = 분자량 * PV / (RT)</p>
<p class=WL1>무게 = 46g/mol * 1atm * 0.001L / (0.08205784 L·atm·K<sup>-1</sup>·mol<sup>-1</sup> * (273+20)K)</p>
<p class=WL1>무게 = 46g *0.001 / (0.08205784*293)</p>
<p class=WL1>무게 = 46 / (82.05784*293) = 0.0019132g = <span style="font-weight:bold; ">1.9132㎎</span></p>
<p class=WL1>완전히 기화된 개미산 1㎎의 부피는 이 값의 역수이므로 0.5227㎖</p>
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<p class=WL1>개미산의 분자량은 46임.</p>
<p class=EL1> </p>
<p class=WL1>1. 20℃, 1기압 하에서 완전히 기화된 개미산 1㎖의 무게는?</p>
<p class=WL1>PV = nRT (∵n=몰수=무게/분자량)</p>
<p class=WL1>n = PV/(RT)</p>
<p class=WL1>무게/분자량 = PV / (RT)</p>
<p class=WL1>무게 = 분자량 * PV / (RT)</p>
<p class=WL1>무게 = 46g/mol * 1atm * 0.001L / (0.08205784 L·atm·K<sup>-1</sup>·mol<sup>-1</sup> * (273+20)K)</p>
<p class=WL1>무게 = 46g *0.001 / (0.08205784*293)</p>
<p class=WL1>무게 = 46 / (82.05784*293) = 0.0019132g = <span style="font-weight:bold; ">1.9132㎎</span></p>
<p class=WL1>완전히 기화된 개미산 1㎎의 부피는 이 값의 역수이므로 0.5227㎖</p>
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카페 게시글
응애(진드기)
기화된 개미산의 무게
비봉
추천 0
조회 194
22.10.29 18:59
댓글 2
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다음검색
첫댓글 비봉님 저는 화학에 대해서 문외한이라서 한가지 문의 합니다.
개미산 처리시에 85% 원액을 60%로 희석할시에 부피 기준이 옳다고 생각합니다만,
혹자는 무게 기준으로 계산해야 한다고 합니다.
물이 액체요,개미산도 고체가 아닌 액체 이므로 부피 기준이 맞다고 봅니다.
개미산의 비중이 1.22이므로 액체,고체기준에따라 비율이 차이가 납니다.
말통으로 판매되는 85% 농도의 개미산은 무게로 85%입니다.
이 것을 부피농도로 환산해보면 85%가 아닙니다.
굳이 부피농도로 희석하려면 무게농도 85%에 해당하는 부피농도가 얼마인지부터 알고 하시면 되겠습니다.
무게농도인지 부피농도인지는 그 형태가 고체인지 액체인지로 나뉘지 않습니다.