Предмет: Химия, автор: belozerovpavel319

Определите массовую долю нового раствора, если к 400г 5% раствора добавили 30г соли. (с условием)​

Ответы

Автор ответа: jyliapw
1

Ответ:

Дано

m(вещ)-400г

w1- 5%

m(раств)-30г

Найти

w2-?

решение

m(вещ) = w1 * m(вещ) : 100 = 5*400 : 100= 20г

m(раств)=m(раств)+m(вещ) = 400+30=430г

m(вещ)1 = m(вещ)+m(раств) = 20+30=50г

W2= m(вещ)1 : m(раств) = 50:430=0.11 або 11%

ответ 11%

Автор ответа: mrvladimir2
1

Ответ: 11,63%

Дано:

m₁(p-pa соли) = 400 г

ω₁(соли) = 5% или 0,05

m₂(соли) = 30 г

ω₂(соли) - ?

Объяснение:

1. Находим массу соли в начальном р-ре:

m₁(соли) = m₁(p-pa соли)*ω₁(соли) = 400 г*0,05 = 20 г

2. Находим массу соли в конечном р-ре:

m(cоли) = m₁(соли) + m₂(соли) = 20 г + 30 г = 50 г

3. Находим массу конечного р-ра:

m₂(p-pa cоли) = m₁(p-pa соли) + m₂(соли) = 400 г + 30 г = 430 г

4. Находим массовую долю нового раствора:

ω₂(соли) = m(cоли)/m₂(p-pa cоли) = 50 г/430 г = 0,1163 или 11,63%

Ответ: 11,63%

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