现有100克含盐18%的盐水,如果要加水配成含盐10%的盐水,需耍加水多少克?✀

2025-03-15 03:42:49
推荐回答(3个)
回答1:

需耍加水80克。

根据题意,现有100克含盐18%的盐水,

运用乘法,列式可得:

100*18%=18克,即这100克的盐水中有18克盐,

那么这杯水中的水有100-18=72克。

要加水配成含盐10%的盐水,

设加X克水,可列式为:

18/(100+X)=10%

解此式可得:X=80

所以需要加水80克。

扩展资料:

此类问题属于数学中的浓度三角问题,此类问题的解题公式为:

浓度=溶质/(溶剂+溶质)

解题思路:把混合的两种物质并排写在一张纸上,把他们的浓度和质量写在他们旁边,把混合后的物质写在那两个物质的中间的下方,这样三种物质形成了一个倒三角形。

比如说进行混合的两物质是A和B,浓度分别是m和n,质量分别是x和y,混合后的物质是C,浓度是p,质量是z。

那么满足如下关系:z=x+y p-m:n-p=y:x。所以m,n,p,x,y,z这六个量只要知道4个其他两个就可以求出了。

参考资料来源:百度百科—浓度三角

回答2:

100×18%÷10%-100
=180-100
=80克

回答3:

80la lo lo

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