i need help with the unknown % calculation

Calculations for standardization: $Mol HCl = \frac{0.238 g Na _{2} CO _{3}}{105.988 g / mol Na _{2} CO _{3}} \times \frac{2 mol HCl}{1 mol Na _{2} CO _{3}} = 0.00449 mol$ $MHCl = \frac{mol HCl}{V HCl yellow endpoint} = \frac{0.00449 mol}{0.0451 L} = 0.0995 M$ Conclusion:

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i need help with the unknown % calculation

Calculations for standardization:

Mol HCl = \frac{0.238 g Na _{2} CO _{3}}{105.988 g / mol Na _{2} CO _{3}} \times \frac{2 mol HCl}{1 mol Na _{2} CO _{3}} = 0.00449 mol

MHCl = \frac{mol HCl}{V HCl yellow endpoint} = \frac{0.00449 mol}{0.0451 L} = 0.0995 M

Conclusion:

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