Đáp án:

 \[\text C\%_{\text H_3\text{PO}_4}=3,344\%\]

Giải thích các bước giải:

 Giả sử có 6 lít dung dịch X

Suy ra: \(\begin{cases}\text V_{\text{dd}\ H_3PO_4\ 6\%}=1\ \text{lít}\\ \text V_{\text{dd}\ H_3PO_4\ 4\%}=2\ \text{lít}\\\text V_{\text{dd}\ H_3PO_4\ 2\%}=3\ \text{lít}\end{cases}\\⇔ \begin{cases}\text m_{\text{dd}\ H_3PO_4\ 6\%}=1000\cdot 1,03=1030\ \text{gam}\\ \text m_{\text{dd} \ H_3PO_4 \ 4\%}=2000\cdot 1,02=2040\ \text{gam}\\ \text m_{\text{dd}\ H_3PO_4\ 2\%}=3000\cdot1,01=3030\ \text{gam}\end{cases}\)

\(⇔ \begin{cases}m_{H_3PO_4(\text{dd}\ H_3PO_4 \ 6\%)}=1030\cdot 6\%=61,8\ \text{gam}\\ m_{H_3PO_4(\text{dd}\ H_3PO_4 \ 4\%)}=2040\cdot 4\%=81,6\ \text{gam}\\ m_{H_3PO_4(\text{dd}\ H_3PO_4 \ 2\%)}=3030\cdot 2\%=60,6\ \text{gam}\end{cases}\Rightarrow \sum m_{H_3PO_4}=61,8+81,6+60,6=204\ \text{gam}\)

Suy ra nồng độ % của axit photphoric là: 

\[\text C\%_{\text H_3\text{PO}_4}=\dfrac{204}{1030+2040+3030}\cdot 100\%=3,344\%\]